Operations with Power Series: Learn It 2

Multiplication of Power Series

You can also create new power series by multiplying two existing power series together. This multiplication technique gives us another powerful method for finding power series representations of functions.

The process works similarly to multiplying polynomials. When you multiply two power series, you collect like terms to form the coefficients of the resulting series.

For example, suppose you want to multiply:

[latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots[/latex]

and

[latex]\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots[/latex].

The product becomes:

[latex]\begin{array}{cc}\hfill \left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}\right)\left(\displaystyle\sum _{n=-0}^{\infty }{d}_{n}{x}^{n}\right)& =\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots \right)\cdot \left({d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots \right)\hfill \\ & ={c}_{0}{d}_{0}+\left({c}_{1}{d}_{0}+{c}_{0}{d}_{1}\right)x+\left({c}_{2}{d}_{0}+{c}_{1}{d}_{1}+{c}_{0}{d}_{2}\right){x}^{2}+\cdots .\hfill \end{array}[/latex]
Notice how each coefficient in the product collects all terms that multiply to give the same power of [latex]x[/latex]. For [latex]x^n[/latex], you get contributions from [latex]c_0 d_n + c_1 d_{n-1} + c_2 d_{n-2} + \cdots + c_n d_0[/latex].

theorem: multiplying power series

If [latex]\sum_{n=0}^{\infty} c_n x^n[/latex] and [latex]\sum_{n=0}^{\infty} d_n x^n[/latex] converge to functions [latex]f[/latex] and [latex]g[/latex] respectively on a common interval [latex]I[/latex], then:

Define the new coefficients as:

[latex]\begin{array}{cc}\hfill {e}_{n}& ={c}_{0}{d}_{n}+{c}_{1}{d}_{n - 1}+{c}_{2}{d}_{n - 2}+\cdots +{c}_{n - 1}{d}_{1}+{c}_{n}{d}_{0}\hfill \\ & ={\displaystyle\sum _{k=0}^{n}{c}_{k}{d}_{n-k}}.\hfill \end{array}[/latex]

 The product series is:

[latex]\left({\displaystyle\sum _{n=0}^{\infty}}{c}_{n}{x}^{n}\right)\left(\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}\right)=\displaystyle\sum _{n=0}^{\infty }{e}_{n}{x}^{n}[/latex]

This new series converges to [latex]f(x) \cdot g(x)[/latex] on interval [latex]I[/latex].

 

This resulting series is called the Cauchy product.

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course.

We can use this multiplication technique to find power series for functions like: [latex]f(x) = \frac{1}{(1-x)(1-x^2)}[/latex] by multiplying the known power series for [latex]y = \frac{1}{1-x}[/latex] and [latex]y = \frac{1}{1-x^2}[/latex].

Multiply the power series representation

[latex]\begin{array}{cc}\hfill \frac{1}{1-x}& ={\displaystyle\sum _{n=0}^{\infty}}{x}^{n}\hfill \\ & =1+x+{x}^{2}+{x}^{3}+\cdots \hfill \end{array}[/latex]

 

for [latex]|x|<1[/latex] with the power series representation

[latex]\begin{array}{cc}\hfill \frac{1}{1-{x}^{2}}& ={\displaystyle\sum _{n=0}^{\infty}}{\left({x}^{2}\right)}^{n}\hfill \\ & =1+{x}^{2}+{x}^{4}+{x}^{6}+\cdots \hfill \end{array}[/latex]

 

for [latex]|x|<1[/latex] to construct a power series for [latex]f\left(x\right)=\frac{1}{\left(1-x\right)\left(1-{x}^{2}\right)}[/latex] on the interval [latex]\left(-1,1\right)[/latex].

Watch the following video to see the worked solution to the above example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.2.3” here (opens in new window).